Life In 19x19

drmwc wrote:I may have phrased the alternative badly.

We bet on a fair coin. If it's tails, you win £1. I it's heads, I win £2 since you are forced to place the wager twice.

What odds do you require?

The second bet is a sure loss for me, and we both know that. To ask what odds I require is ridiculous. You are forcing me to give 2:1 odds on a 50:50 bet. (Or you are forcing me to make a sure loss bet if the coin comes up tails. Either way you look at it, I am the one giving odds.)

Edit: To be clear. I want 50:50 odds on the bet as a whole, instead of my laying 2:1. Or, if we are separating the bets, I want to bet 0 on the sure loss bet, with 50:50 odds on the coin toss bet.

I have a question about the first puzzle. Why have a box with two silver coins? It doesn't add anything right?

BigBadBuu wrote:I have a question about the first puzzle. Why have a box with two silver coins? It doesn't add anything right?

It does. Each of the 3 boxes matters. See Post #50.

EdLee wrote:

BigBadBuu wrote:I have a question about the first puzzle. Why have a box with two silver coins? It doesn't add anything right?

It does. Each of the 3 boxes matters. See Post #50.

Yeah what the hell was that? English dude english

BigBadBuu wrote:Yeah what the hell was that? English dude english

That was no higher than 9th grade English, and between 4th and 9th grade math,
depending on which country's math education.
"increment" here meant "add 1 to"; "iterations" here meant "repetitions."

Let's label the 3 bowls:
A:
B:
C:

If we remove B, then the first bowl randomly picked has a 100% chance of it being ( either A or C ).
If we keep B, then the first bowl randomly picked has only a 2/3 chance of it being ( either A or C ).

So whether B is there or not affects the outcome of the first bowl picked, which also affects the outcome of the second stone.

EdLee wrote:

BigBadBuu wrote:Yeah what the hell was that? English dude english

That was no higher than 9th grade English, and between 4th and 9th grade math,
depending on which country's math education.
"increment" here meant "add 1 to"; "iterations" here meant "repetitions."

Let's label the 3 bowls:
A:
B:
C:

If we remove B, then the first bowl randomly picked has a 100% chance of it being ( either A or C ).
If we keep B, then the first bowl randomly picked has only a 2/3 chance of it being ( either A or C ).

So whether B is there or not affects the outcome of the first bowl picked, which also affects the outcome of the second stone.

in fact here removing the bowl WILL give the same result:


applying Bayes therorem on the 3 bowl problem:
P(we picked bowl A | the extracted stone was ) = P(we picked a stone | A).P(A)/P(we picked a stone) = 1. 1/3/(1/2) = 2/3.
because: P(A)=1/3 (obvious)
P(we picked a stone | A) = 1 (A contains only B stones)
P(we picked a stone) = 1*P(A)+1/2*P(B)+0.P(C)=1/3+1/2*1/3=3/6=1/2
Now with only bowls A and B:

P(we picked bowl A | the extracted stone was ) = P( | A).P(A)/P( ) = 1. 1/2/(3/4) = 2/3. !!!
because:
P(A)=1/2 (obvious)
P(we picked a stone | A) = 1 (A contains only B stones)
P(we picked a stone) = 0.5*P(A)+1/2*P(B)+0.P(C)=1/2+1/4=3/4

different computation, same result

If we try to change the numbe of stones in each bowl, we can see what is needed for the 2 problems to give the same answer:

we need tha P( in the 3 bowl problem)/P( in the 2 bowl problem)=2/3. (ie the proba of picking a stone must cancel out the change of proba in picking bowl A whne we go from 2 bowl to 3 bowls)

This is true as long as there are as much stones in the A and the B bowl, and if the bowl B contains as many and stones.

The number of stones in the bowl C is irrelevant (which make sense as if we pick a stone we know we did not pick it)!!

EdLee wrote:So whether B is there or not affects the outcome of the first bowl picked, which also affects the outcome of the second stone.

As perceval already pointed out, it doesn't. What it affects is the probability of picking a white (or black stone) initially, but not the probability being asked, which was (roughly) "how likely is it that the second stone in the same bowl will be of the same color". If you first pick a white stone, it doesn't matter how many bowls with only black stones there were to begin with (or, if you pick a black stone, it doesn't matter how many bowls with only white stones there were to begin with).

One thing that surprised me about the Sleeping Beauty problem is the claim that the Bayesians think that Sleeping Beauty should believe that the probability that the coin came up heads is 1/2 , while the frequentists think that she should believe that the probability is 1/3. I don't believe that there should be any difference, but if there was one, I would have thought that it would be the other way around. After all, the frequency of heads is 1/2, and the application of Bayes Theorem gives you a probability of 1/3. Frankly, I am gobsmacked.

Here is a simplification of the problem that illustrates the main point. A fair coin is tossed. If it comes up tails, Sleeping Beauty is asked what is the probability that it came up heads. If it comes up heads, she is not asked. Sleeping Beauty knows all of this. What should her answer be?

Well, duh! This is not even worth hiding. It is hardly worth stating. The probability that the coin came up heads is 0.

But why is that? The key answer is that for Sleeping Beauty the fact that she is being asked what the probability is is evidence about the probability. (That is unusual in real life, which is part of what makes this a problem. ) By Bayes Theorem the posterior odds that the coin came up heads versus tails is the product of the prior odds of the same times the odds that Sleeping Beauty will be asked about the probability if the coin came up heads versus that she will be asked if the coin came up tails. The prior odds are 1, the odds about being asked are 0, and the posterior odds are 0. That makes the probability 0. Bingo!

In the more complex problem Sleeping Beauty's amnesia guarantees that the conditions of each question are the same. (Again, not like real life. ) We condition on what she knows, not on what has happened. Thus Sleeping Beauty's answer should be the same for each question. She is asked once if the coin came up heads and twice if it came up tails. So the odds that she will be asked if the coin came up heads versus if it came up tails are 1:2. By Bayes Theorem, the posterior odds that the coin came up heads versus tails are also 1:2, and so the probability that the coin came up heads is 1/3. Bingo, again.

tj86430 wrote:As perceval already pointed out, it doesn't. What it affects is the probability of picking a white (or black stone) initially, but not the probability being asked, which was (roughly) "how likely is it that the second stone in the same bowl will be of the same color". If you first pick a white stone, it doesn't matter how many bowls with only black stones there were to begin with (or, if you pick a black stone, it doesn't matter how many bowls with only white stones there were to begin with).

perceval and TJ, thanks --

Two simulations:

Simulation 1:
(0) 3 bowls: ( ), ( ), and ( ).
(0a) Set X = 0.
(1) Randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) This stone turns out to be .
(3a) IF second stone in the bowl is also , increment X.
(4) Go back to (1); repeat this 1 million times. What is the approx. value of X after 1 million iterations ?

Simulation 2:
(0) 102 bowls: ( ) and ( ); plus 100 ( ).
(0a) Set Y = 0.
(1) Randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) This stone turns out to be .
(3a) IF second stone in the bowl is also , increment Y.
(4) Go back to (1); repeat this 1 million times. What is the approx. value of Y after 1 million iterations ?

Re: Bill's "pragmatics" -- I apologize because I'm easily confused by the wording of "What's the probability of..." --
could someone run the above two (simple) simulations?
I'd like to know the values of X and Y. Thanks.

I'd also like to know, in the original question: " What is the probability that the second stone in the bowl is also ? ",
is it asking what is the value of (X/N) as N tends to infinity ? (where N is the number of interations.)

And, to understand BigBadBuu's quesiton, will (X = Y) as N tends to infinity ?

EdLee wrote:

tj86430 wrote:As perceval already pointed out, it doesn't. What it affects is the probability of picking a white (or black stone) initially, but not the probability being asked, which was (roughly) "how likely is it that the second stone in the same bowl will be of the same color". If you first pick a white stone, it doesn't matter how many bowls with only black stones there were to begin with (or, if you pick a black stone, it doesn't matter how many bowls with only white stones there were to begin with).

perceval and TJ, thanks --

Two simulations:

Simulation 1:
(0) 3 bowls: ( ), ( ), and ( ).
(0a) Set X = 0.
(1) Randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) This stone turns out to be .
(3a) IF second stone in the bowl is also , increment X.
(4) Go back to (1); repeat this 1 million times. What is the approx. value of X?

Do you mean for (2) If this stone turns out to be .

In that case you are asking how often you pick the bowls with .

Or do you want to ask, given that the chosen stone is , how often is the other stone ? If so, you need to count how often the chosen stone is .

E. g.:

(0) 3 bowls: ( ), ( ), and ( ).
(0a) Set X = 0.
(0b) Set Z = 0.
(1) Randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) If this stone turns out to be , increment Z.
(3a) IF second stone in the bowl is also , increment X.
(4) Go back to (1); repeat this until Z = 1 million. What is the approx. value of X?

Simulation 2:
(0) 102 bowls: ( ) and ( ); plus 100 ( ).
(0a) Set Y = 0.
(1) Randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) This stone turns out to be .
(3a) IF second stone in the bowl is also , increment Y.
(4) Go back to (1); repeat this 1 million times. What is the approx. value of Y after 1 million iterations ?

Same deal. Do you want to know how often you pick the bowl? Or do you want to know how often the second stone is , given that the chosen stone is ?

Hi Bill,

Bill Spight wrote:Do you mean for (2) If this stone turns out to be .
In that case you are asking how often you pick the bowls with .

Thanks; yes -- I used the same wording as perceval's original.

Bill Spight wrote:Or do you want to ask, given that the chosen stone is , how often is the other stone ? If so, you need to count how often the chosen stone is .

(0) 3 bowls: ( ), ( ), and ( ).
(0a) Set X = 0.
(0b) Set Z = 0.
(1) Randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) If this stone turns out to be , increment Z.
(3a) IF second stone in the bowl is also , increment X.
(4) Go back to (1); repeat this until Z = 1 million. What is the approx. value of X?

Also correct -- it's not me who's asking --
this is precisely why I found the original wording to be confusing:

This stone turns out to be . What is the probability that the second stone in the bowl is also ?

Is the original question asking for (X/N) or (X/Z) ? We know (X/N) = 1/3 as N -> Inf.
We also know (X/Z) = 1/2 as N -> inf. Which is why some people thought 50% to the original question.
Or, maybe the original question is asking for another ratio -- what is it?
And how are we supposed to determine which ratio the original question is asking for ?
Maybe the original wording is 100% unambiguous in the field of probability,
but it's been a few years since school, so I'm probably^(*) very much out of it.
This is why I took the little green alien bit (which you replied to). Thanks.

^(*) Stolen from Tom Stoppard.

EdLee wrote:Hi Bill,

Bill Spight wrote:Do you mean for (2) If this stone turns out to be .
In that case you are asking how often you pick the bowls with .

Thanks; yes -- I used the same wording as perceval's original.

Bill Spight wrote:Or do you want to ask, given that the chosen stone is , how often is the other stone ? If so, you need to count how often the chosen stone is .

(0) 3 bowls: ( ), ( ), and ( ).
(0a) Set X = 0.
(0b) Set Z = 0.
(1) Randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) If this stone turns out to be , increment Z.
(3a) IF second stone in the bowl is also , increment X.
(4) Go back to (1); repeat this until Z = 1 million. What is the approx. value of X?

Also correct -- it's not me who's asking --
this is precisely why I found the original wording to be confusing:

This stone turns out to be . What is the probability that the second stone in the bowl is also ?

Is the original question asking for (X/N) or (X/Z) ? We know (X/N) = 1/3 as N -> Inf.
We also know (X/Z) = 1/2 as N -> inf. Which is why some people thought 50% to the original question.

Do we know that?

Bill, thanks, I'll have to think about (X/Z). But the main thing I was confused about
was whether the original question was asking for (X/Z) (versus X/N). Thanks!

The key issue with the sleeping beauty problem is that SB has no way to differentiate Monday and Tuesday. This leads to some conterinuitive features.

Consider the statement "Given that the coin landed head, what is the probability that today is Tuesday?" Intuitively, the answer is 1/2. However, this leads to problems.

Formally, we have the following statements we wish to be true:

(1) P(Mon or Tue) = 1
(2) P(heads and Tue) = 0
(3) P(heads | Mon or Tue) = 1/2
(4) P(heads | Mon) = 1/2
(5) P(Mon | tails) = 1/2

We know that 1 is true from the set up of the experiment.

Similarly. we know that 2 is also true.

We want 3 to be true because we are told that the coin is fair (and presumably SB verified this before the experiment.) Also, SB gains no information from waking.

We want 4 to be true, again because the coin is fair and SB gains no new information from waking on Monday.

We want 5 to be true for reasons such as the indfference principle.

However, the issue is that 1-5 above are inconsistent - they do not define a valid probability measure. (I've left the proof as an exercise to the reader.)

People who insist that the answer is 1/3 reject hypothesis 3 above, and insist that P(heads | Mon or Tue) = 1/3. This does give a consistent probability measure. However, changing the probabilty of heads when no new information is available is offensive to me, and feels hard to justify.

The neatest resolution is that 5) above has no defined probability. If, at the interview, SB is asked "What is the probability today is Monday?", her reply is: "I do not have sufficient information to answer".

The key issue with the question is the use of the word "today". This means the statement uses an indexical, and its truth value changes dependent on circumstances. Other indexicals include "I", "now", "this" etc.

Probability behaves badly when we attempt to assign probabilities to statements which include indexicals. Finding a replacement for the statement "Today is Monday" without fundamentally changing the problem is difficult - I have not seen it done.

For example, suppose an inspector arrvies either on Monday or Tuesday and attends the interview (if there is one). The day she arrives is chosen at random with probability 1/2, and is independent of the coin flip.

Now the symmetry of Monday and Tuesday collapses, and frequentists and Bayesians would both agree that probability of heads, given the inspector is present at the interview when asked, is 1/3. But this seems to violate the spirit of the oringinal experiment, where SB could not differentiate Monday and Tuesday.

drmwc wrote:The key issue with the sleeping beauty problem is that SB has no way to differentiate Monday and Tuesday. This leads to some conterinuitive features.

Consider the statement "Given that the coin landed head, what is the probability that today is Tuesday?" Intuitively, the answer is 1/2. However, this leads to problems.

Wait a second. Which version is this about? If it is the one in which Sleeping Beauty is awakened on Tuesday if and only if the coin came up tails, then intuitively, the answer is 0.

Formally, we have the following statements we wish to be true:

Starting with what we wish to be true does not always work.

(1) P(Mon or Tue) = 1
(2) P(heads and Tue) = 0
(3) P(heads | Mon or Tue) = 1/2
(4) P(heads | Mon) = 1/2
(5) P(Mon | tails) = 1/2

I suppose you mean the conditions to be what Beauty knows.

We know that 1 is true from the set up of the experiment.

Similarly. we know that 2 is also true.

We want 3 to be true because we are told that the coin is fair (and presumably SB verified this before the experiment.) Also, SB gains no information from waking.

Of course Beauty gains information from waking. The probability that she wakes up on Monday is 1. The probability that she wakes up on Tuesday is 1/2. So when she wakes up, the odds are 2:1 that it is Monday. That's information.

Now, you may say that Beauty already had that figured out on Sunday. But on Sunday the probability that it was Monday was 0. It is easy to go astray with such statements as those about gaining or losing information when waking. The question is what are the conditions of the probabilities. On Sunday, the probability was 1/2 that the coin would come up heads. On Monday and Tuesday, the probability that it had come up heads does not have to be the same as it was on Sunday, because the conditions are different.

We want 4 to be true, again because the coin is fair and SB gains no new information from waking on Monday.

What information Beauty gains from waking on Monday is not the question. The point is that Beauty always wakes up on Monday, regardless of the result of the coin toss. But on awakening, Beauty is not sure that it is Monday. She has to be told in order to get the "Mon" condition for 4.

We want 5 to be true for reasons such as the indfference principle.

That is one reason that probability is empirical.

However, the issue is that 1-5 above are inconsistent - they do not define a valid probability measure. (I've left the proof as an exercise to the reader.)

People who insist that the answer is 1/3 reject hypothesis 3 above, and insist that P(heads | Mon or Tue) = 1/3. This does give a consistent probability measure. However, changing the probabilty of heads when no new information is available is offensive to me, and feels hard to justify.

The "no new information" phrase is a trap. The right way to think about these things is in terms of the conditions of the probabilities. True, in one interpretation of "no new information", Sleeping Beauty gains no new information. But the conditions of the probabilities are different, as indicated above. That is what is important.

The neatest resolution is that 5) above has no defined probability. If, at the interview, SB is asked "What is the probability today is Monday?", her reply is: "I do not have sufficient information to answer".

The key issue with the question is the use of the word "today". This means the statement uses an indexical, and its truth value changes dependent on circumstances. Other indexicals include "I", "now", "this" etc.

Probability behaves badly when we attempt to assign probabilities to statements which include indexicals. Finding a replacement for the statement "Today is Monday" without fundamentally changing the problem is difficult - I have not seen it done.

For example, suppose an inspector arrvies either on Monday or Tuesday and attends the interview (if there is one). The day she arrives is chosen at random with probability 1/2, and is independent of the coin flip.

Now the symmetry of Monday and Tuesday collapses, and frequentists and Bayesians would both agree that probability of heads, given the inspector is present at the interview when asked, is 1/3. But this seems to violate the spirit of the oringinal experiment, where SB could not differentiate Monday and Tuesday.

Since Beauty cannot tell the difference between Monday and Tuesday, we should regard them as simply labels for different cases. I don't think that there is a problem with "today", meaning "this case", any more than there is a problem, if we draw two cards from a well shuffled standard deck and leave them face down, of then pointing to one card and asking what is the probability that this card is a Jack.

But Monday and Tuesday are not symmetrical, anyway. Beauty just does not know which is which.

----

Here is a variant that restores the symmetry between Monday and Tuesday as far as Beauty's awakening goes, but retains the asymmetry of the original problem. I'll skip the setup.

On Monday after Beauty awakes she is asked what is the probability that the coin came up heads. She is given the amnesia drug to make her forget the events of Monday, but she is allowed to awaken on Tuesday.

On Tuesday after Beauty awakes, if the coin came up tails she is asked what is the probability that it came up heads. If it came up heads, she is not asked anything about how the coin came up.

Upon awakening, what probability should Beauty assign to the coin coming up heads? Plainly, 1/2. When asked, what probability should she assign? Plainly -- I trust --, 1/3.

In this variation awakening does not affect her prior belief in the odds that the coin came up heads, but being asked (by the experimenters) does. She is more likely to be asked when the coin came up tails. When Beauty wakes up, Monday and Tuesday are symmetrical. It is the asking that breaks the symmetry.

If we eliminate the case in which Beauty wakes up but is not asked about how the coin came up, the symmetry is again destroyed.